Imagine that you are in chemistry lab and need to make 1.00 L of a solution with
ID: 519281 • Letter: I
Question
Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.80.
You have in front of you
100 mL of 6.00×102M HCl,
100 mL of 5.00×102M NaOH, and
plenty of distilled water.
You start to add HCl to a beaker of water when someone asks you a question. When you return to your dilution, you accidentally grab the wrong cylinder and add some NaOH. Once you realize your error, you assess the situation. You have 84.0 mL of HCl and 90.0 mL of NaOHleft in their original containers.
Assuming the final solution will be diluted to 1.00 L , how much more HCl should you add to achieve the desired pH?
Express your answer to three significant figures and include the appropriate units.
Explanation / Answer
pH = 2.80
pH = - log [H+]
-log[H+] = 2.80
[H+] = 1×10^-2.80
= 0.00158M
volume of solution needed = 1Litre
Mol of HCl needed = 0.00158
[HCl] = 0.06M
Volume of HCl added = 100ml - 84ml = 16ml
No of mol of HCl in 16ml = (0.06mol/1000ml)×16ml = 0.00096
[NaOH] = 0.05M
Volume of NaOH added = 100ml - 90ml = 10ml
No of mol of NaOH in 10ml = 0.0.0005
0.0005 mol of NaOH react with 0.0005mol HCl
Remaining HCl mol = 0.00096-0.0005 = 0.00046
Mole of HCl further to add = 0.00158 - 0.00046 = 0.00112
Volume of HCl having 0.00112 mol = (1000ml/0.060mol)×0.00112mol =18.67ml
Therefore, 18.67ml of HCl solution should be added more to the beaker.